Everyone, in earlier videos, we saw how to solve PV diagram problems. And in more recent videos, we've looked at heat engine problems. Well, some problems like the one that we're going to work out below will give you a PV diagram for a heat engine and then ask you to calculate something, like the total work done by the engines. We're going to put those ideas together in this video, and I'm going to show you how to calculate the work in heat engines by using the PV diagram. And what we're going to see is we're going to see the relationship between the PV diagram steps and the quantities that we use in our energy flow diagrams. So let's go ahead and get started. We're actually just going to jump straight into our problem because there's a lot of stuff that we already know how to do. So let's check it out. In this problem, we have 2 moles of a gas, and it's basically undergoing a cyclic process. What we're told here is that heat is removed from this gas at constant pressure from a to b. Remember, that just means that it's an isobaric process. Then from b to c, we're adding heat at constant volume. So this is isovolumetric. And then finally, the gas is going to expand isothermally. So that just means it's going to sort of ride along this curve right here, go back to a, and then the whole thing is going to repeat over again. So that's actually the first point I want to make here. Remember that heat engines are cyclic processes. They start and end at the same point at the same initial state. But what you need to know here is that on PV diagrams, they will actually always run-in clockwise loops. Notice how we've gone through a cycle like this, but the overall pattern of this loop is in the clockwise direction. Alright. So that's always going to be true for heat engines. Alright. So let's take a look at this problem here, because the first thing we want to calculate in part a is the heat transfer of each process. So we have 3 processes we want to calculate q for. So we have q from a to b, and then we have q from b to c, and then we have q from c back to a again. Alright? So we spent a lot of time developing our equations in this table here for special processes. That's the that's basically what we're going to do here. We're just going to look at each process and then figure out which equation of q that we use from this row here. So let's get started. So from a to b, we know this is an isobaric process, constant pressure. So we look th...

We have 2 moles. CP just comes from this table here. It's a monoatomic gas, so we're going to use \( \frac{5}{2} R = \frac{5}{2} \times 8.314 \). Now what's the change in temperature? We're going between 2 isotherms, 200 back to 100. So this is basically just going to be \( \Delta T = -100 \). So you go ahead and work this out. What you're going to get is -4157 joules. Now for the second one, we're going to use a similar process. \( Q_{bc} \) is equal to, this is an isovolumetric process. So we're going to use this equation over here. Very much the same process. We're just going to use a slightly different molar specific heat. This is going to be 2. Now we're not going to use \( \frac{5}{2} \). We're going to use \( \frac{3}{2} \) for monoatomic. Right? So this is going to be \( \frac{3}{2} \times 8.314 \). And now what's the delta t? Well, here we're going from 100 back to 200 again. So this is going to be \( \Delta T = 100 \). So we're going to use 100 here, and what you're going to get is 2494 joules.

Finally, this last process here, \( q \) from c to a, is an isothermal one. So we look through our equations and \( q = w \). Whatever the work done in this process is, that's also the heat transfer. Now do we use this equation then? Well, actually, we don't need to because in this problem, we're told that this process here or this gas does 23100 joules of work. So this is \( q = w = 23100 \) joules, and there's no calculation necessary. Alright? So that's all there is to it. Right? That's just all the heat transfers using a bunch of equations we've seen before. So let's now go on to part b. We want to calculate the work done by the heat engine.

So that's going to be the work done by the engine. Now how do we do this? Well, you may remember from our video on cyclic processes that whenever you have a cycle, the work that's done is going to be the area that is enclosed within the loop. So in other words, the work done is going to be the area. So how do we do that? Well, unfortunately, there's a couple of problems here because, 1, we have this sort of curved isothermal process, so we can't use something like a triangle or anything like that. And we also don't have any of the values for pressure or volume or anything that we need to calculate the area. So while the work done by the engine is actually the area that's in this shape, we just can't calculate it. So we're going to have to use a different equation. Now in more recent videos, remember, we've talked a lot about heat engines, and we've been using these energy flow diagrams. And remember, for heat engines, the work done is just equal to \( q_h - q_c \). Right? In a heat engine, some heat comes in from the hot reservoir. The engine produces some usable energy work and then spits out the waste heat to the cold reservoir. And the work done is just the q in minus the q out. So we just have to find those. So how do we do that? Which numbers are going to be are they going to be? Or is it going to be this 1? Is it going to be this 1 or this 1? The answer is we're going to actually have to use all of them. So remember that we're going to use \( q_h - q_c \), and \( q_h \) is really just the heat that's added into the system. Remember, we have we have heat added into the system from the hot

Now we do the same exact thing for the \( q_c \). In this problem, we calculated one value of \( q \) that was negative. In this key in this process, what happens is heat is flowing out of the system, and it's flowing to the cold reservoir. So \( q_c \) is the heat removed out of the system to the cold reservoir, and it's going to be the sum of all negative \( q \)'s over the cycle. So this is just going to be -4157. Now just be very careful here because whenever you use these energy flow diagrams, we always want these to be positive numbers, so you may see a bunch of absolute value signs. So basically, what's happening is that this negative sign when you calculate it just means the heat is being removed, but when we use it in our energy flow diagrams, we just use it as a positive number. So you may just see these written with a bunch of absolute value signs. Alright? So that just means that the work done by the engine is just, 4794 minus 4157 because we're using the absolute value here. Alright? Just be really careful with the signs here. You don't want to subtract this because this already has a negative sign, and if you mess this up, you're going to get the wrong answer. Right? So just be very careful. Make sure you're subtracting them, and you're going to get 637 joules. Alright? So that's it for this one, guys. Let me know if you have any questions.